Used
for comparing frequencies (counts) of nominal or ordinal level data for two samples across
two or more subgroups displayed in a crosstabulation table. More
common and more flexible than ztests of proportions.
Problem:
You wish to evaluate the association
between a person's sex and their attitudes toward school spending on athletic programs. A
random sample of adults in your school district produced the following table (counts).

Female 
Male 
Row
Total 
Spend
more money 
15 
25 
40 
Spend
the same 
5 
15 
20 
Spend
less money 
35 
10 
45 
Column
Total 
55 
50 
105 
Assumptions
Independent
random sampling
Nominal/Ordinal
level data
No
more than 20% of the cells have an expected frequency less than 5
No
empty cells
State
the Hypothesis
Ho:
There is no association
between a person's sex and their attitudes toward spending on athletic programs.
Ha:
There is an association
between a person's sex and their attitudes toward spending on athletic programs.
Set
the Rejection Criteria
Determine
degrees of freedom df=(# of rows  1)(# of columns  1)
df=(3
 1)(2  1) or df=2
Establish
the confidence level (.05, .01, etc.); Alpha
= .05
Based
on the chisquare distribution table, the critical value = 5.991
Compute
the Test Statistic
where
Fo=
observed frequency
Fe=
expected frequency for each cell
Fe=(frequency
for the column)(frequency for the row)/n
Frequency
Observed

Female 
Male 
Row
Total 
Spend
more 
15 
25 
40 
Spend
same 
5 
15 
20 
Spend
less 
35 
10 
45 
Column
Total 
55 
50 
105 
Frequency
Expected

Female 
Male 
Row
Total 
Spend
more 
55*40/105
= 20.952 
50*40/105
= 19.048 
40 
Spend
same 
55*20/105
= 10.476 
50*20/105
= 9.524 
20 
Spend
less 
55*45/105
= 23.571 
50*45/105
= 21.429 
45 
Column
Total 
55 
50 
105 
Chisquare
Calculations

Female 
Male 
Spend
more 
(1520.952)2/20.952 
(2519.048)2/19.048 
Spend
same 
(510.476)2/10.476 
(159.524)2/9.524 
Spend
less 
(3523.571)2/23.571 
(1021.429)2/21.429 
Chisquare

Female 
Male 
Spend
more 
1.691 
1.860 
Spend
same 
2.862 
3.149 
Spend
less 
5.542 
6.096 

21.200 

Decide
Results of Null Hypothesis
Since
the chisquare test statistic 21.2 exceeds the critical value of 5.991, you may conclude
there is a statistically significant association
between a person's sex and their attitudes toward spending on athletic programs. As is
apparent in the contingency table, males are more likely to support spending than females.
Software Output Example
